An arrangement of three parallel straight wires placed perpendicular to the plane of paper carry the same curr

Question

An arrangement of three parallel straight wires placed perpendicular to the plane of paper carry the same current $I$ along the same direction as shown in the figure. The magnitude of force per unit length on the middle wire $B$ is given by:

Accepted Answer

1. **Identify the Physical Law:** The magnetic force per unit length between two long, parallel current-carrying conductors separated by a distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi d}$ .
2. **Analyze the Geometry:** In the standard configuration for this problem (NEET 2017), wire $B$ is at the right-angle vertex of a right-angled isosceles triangle, with wires $A$ and $C$ at the other two vertices, both at a distance $d$ from $B$.
3. **Forces on Wire B:** 
   - Wire $A$ exerts an attractive force per unit length $f_1 = \frac{\mu_0 I^2}{2\pi d}$ on $B$ (since the currents are in the same direction) .
   - Wire $C$ exerts an attractive force per unit length $f_2 = \frac{\mu_0 I^2}{2\pi d}$ on $B$ .
4. **Vector Summation:** Since the wires are arranged such that $AB$ is perpendicular to $BC$, the force vectors $\mathbf{f_1}$ and $\mathbf{f_2}$ are mutually perpendicular. The magnitude of the resultant force per unit length is:
   $$F_{net} = \sqrt{f_1^2 + f_2^2} = \sqrt{2} 	imes f = \sqrt{2} \frac{\mu_0 I^2}{2\pi d}$$
5. **Simplification:** $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$, so the magnitude is $\frac{\mu_0 I^2}{\sqrt{2}\pi d}$.

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